Roots of a polynomial with variables

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Lewis Fer
Lewis Fer el 15 de Jun. de 2021
Comentada: Lewis Fer el 16 de Jun. de 2021
For some problems, we have to to study some notions of stablility and zero polynomials in two variables, my que'stion how we can find the roots or zero polynomials in two variables. for example:
P(x,y)=3*xy -5y^2+7*x^2y
or a nother polynom

Respuestas (2)

Sulaymon Eshkabilov
Sulaymon Eshkabilov el 15 de Jun. de 2021
One of the viable ways to solve such polynomial type equations is to setp up the solution space within which you are seeking the roots to compute and solve them using fzero(). E.g.:
x=linspace(-2,2): % Choose the necessary solution space
for t=1:100
EQ= @(y)(3*x(t)*y-5*y.^2+7*(x(t)^2)*y);
y_roots = fzero(EQ,0);
end
  1 comentario
Lewis Fer
Lewis Fer el 15 de Jun. de 2021
but this method in reality doesn't making a difference to search and find th pole of some matrices like transfer function in two dimensional case

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Paul
Paul el 15 de Jun. de 2021
Don't know the scope of the actual problems of interest, but for the two examples in the question:
syms x y
sol = solve(3*x*y - 5*y^2 + 7*x^2*y == 0,[x y],'ReturnConditions',true);
[sol.x sol.y sol.conditions]
ans = 
syms z1 z2
sol = solve(1 - z1*z2 - 1/2*z1^2 - 1/2*z2^2 + z1^2*z2^2 == 0,[z1 z2],'ReturnConditions',true);
[sol.z1 sol.z2 sol.conditions]
ans = 
  5 comentarios
Paul
Paul el 15 de Jun. de 2021
Apparently there are many solutions to this problem, i.e., many pairs (p,s) that make the determinant equal to zero. The pair (p,s) can be expressed as ( (z+2)/(z-4) , z) for z any number not equal to four. Check
A = [1 2;3 4];
p = @(z)((z+2)./(z-4));
s = @(z)(z);
z = 1;
det(diag([p(z) s(z)]) - A)
ans = -3.3307e-16
z = 8;
det(diag([p(z) s(z)]) - A)
ans = 0
z = 1 + 1i;
det(diag([p(z) s(z)]) - A)
ans = -1.3323e-15 + 3.3307e-16i
Lewis Fer
Lewis Fer el 16 de Jun. de 2021
thank', Paul for your answers your idea help me for my work.

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